Same roots, same polynomial? How to prove characteristic polynomial of $AB = BA$?

Question

I'm giving a (simple) proof that the characteristic polynomial of $AB$ = characteristic polynomial of $BA$ (without using the fact that $AB$ and $BA$ are similar).

$det(AB) = det(A)det(B) = det(B)det(A) = det(BA)$. The determinants of $AB$ and $BA$ are equal. This entails that the roots of the characteristic polynomial of $AB = BA$. Can you infer from this that if the roots of a polynomials $p$a are $q$ are the same, then $p=q$?

If not, then I'm stuck on this proof, and some help would be appreciated.

Edit: It's official. I'm stuck on this proof. How do you prove this?

Answer 1

This is not true in general. Knowing the roots of two polynomials are the same is not enough to conclude they are identical. Indeed, consider

$$p(X) = (1+X)(1-X)$$

and

$$q(X) = \frac{1}{2}(1+X)(1-X)$$

over $\mathbb{Q}$, say. These polynomials have the same roots, but $p \neq q$.

update: (thank you @quid for pointing this out)

Since you are working with characteristic polynomials we do have additional information, namely concerning the leading coefficient of each polynomial and that they factor completely as products of linear terms by definition. With this in mind, knowing the leading coefficients and roots of two characteristic polynomials (and the multiplicity of each root, which is absolutely necessary!) one can conclude that they polynomials are in fact equal.

Answer 2

If $\det M=\det N$, all you can say is that the product of the eigenvalues of $M$ is the same as the product of the eigenvalues of $N$. For example, you could have the eigenvalues $2,1/2$ for $M$ and the eigenvalues $3,1/3$ for $N$.